Allocate memory to variable of type int *a[13][13]

Question

I came across an old code where I need to allocate memory for an array of matrices. Currently it is done by creating a structure matrix and allocating memory by making an arrray of structures.

Like so.

struct matrix
{
    int x[13][13];
};
int main()
{
    matrix *push = (matrix*)malloc(sizeof(matrix) * 1000);
    //do stuff
    free(push);
    return 0;
}

But now the question.. Do I really need to do this? I know I am allowed to declare a variable of type int *matrix[13][13];

But I cant seem to figure out a way to allocate memory to this variable.

I tried matrix = (int***)malloc(sizeof(int[13][13])*1000); which results in E0137 expression must be a modifiable lvalue and honestly in doesnt seem correct either

I know there are many better and creative ways of allocating an array like this, but I am curious

Questions

1. int *matrix[13][13] what will this kind of variable even mean or represent? Is it a pointer to a 2D matrix or s it an pointer to an array of 2D matrices?
2. How will I use the above mentioned variable?
3. How will I allocate memory for the above variable without using any more variables?

Clarifications From comments it seems int *matrix[13][13] represents a matrix of 13x13 int*

What I am asking is actually, is there a way to have a variable with a dynamically allocated first dimension and static 2nd and 3rd dimensions without using structures typedefs etc

EDIT As pointed out by felix what I am looking for is int (*matrix)[13][13] = (int(*)[13][13])malloc(1000*sizeof(matrix));

Answer 1

No, you don't want to use

int *matrix[13][13];

That will declare a [13][13] array of int*s.

You need

// pointer to "an array of 13 x 13 ints"
int (*matrix)[13][13];

or simplify it by using a type alias.

typedef int matrix_type[13][13];
matrix_type* matrix_ptr;

You can allocate memory for such a variable using

matrix_type* matrix_ptr = malloc(1000*sizeof(*matrix_ptr));

and assign values to it using

matrix_ptr[0][0][0] = 0;

...

matrix_ptr[999][12][12] = 0;

Answer 2

I have noticed that sometimes beginner programmers are finding typedef-ing the multidimensional matrices which have to be dynamically allocated easier as it reduces the problem to the single star pointer and the sizeof of the defined type and pointer to it is much easier to understand.

#include <stdio.h>

typedef int my13x13matrix[13][13];

int main(void) {
    my13x13matrix *mylargematrix = malloc(1000 * sizeof(*mylargematrix));

    mylargematrix[999][12][12] = 5;

    printf("%d \n", mylargematrix[999][12][12]);
    return 0;
}

you can also use the single star pointer arithmetic.

my13x13matrix *anotherpointer = &mylargematrix[10];
(*anotherpointer)[5][5] = 1;
anotherpointer++;
(*anotherpointer)[5][5] = 2;

printf("%d %d\n", mylargematrix[10][5][5], mylargematrix[11][5][5]);